INSTITUTE OF MATHEMATICS POLISH ACADEMY OF SCIENCES
WARSZAWA 1992
SOME PROPERTIES OF EIGENFUNCTIONS OF LINEAR PENCILS AND APPLICATIONS TO MIXED TYPE OPERATOR-DIFFERENTIAL
EQUATIONS
S. G. P Y A T K O V
Institute of Mathematics, Russian Academy of Sciences, Siberian Branch Universitetski˘ı Prosp. 4, 630090 Novosibirsk, Russia
Abstract. In the first part of the paper we study some properties of eigenelements of linear selfadjoint pencils Lu = λBu. In the second part we apply these results to the investigation of some boundary value problems for mixed type second order operator-differential equations.
In this paper we consider linear pencils of the form
(1) Lu = λBu
where B, L are selfadjoint operators in a complex separable Hilbert space E.
We suppose that L is a positive operator, though almost the same results can be obtained in the case when the maximal nonpositive L-invariant subspace is finite-dimensional. Let D(L) ⊂ E be the domain of L and let H1 be the com- pletion of D(L) with respect to the norm kukH1 = (Lu, u)1/2. We suppose that D(L) ⊂ D(B) and that there are dense injections of H1 in E and D(|B|1/2). We define F1= (H1∩ ker B)⊥ (the orthogonal complement in H1) and we let F0 be the completion of F1with respect to the norm kukF0 = k |B|1/2ukE. The problem is whether or not the eigenfunctions of the problem (1) constitute a Riesz (un- conditional) basis in F0. The definition and some properties of Riesz bases can be found in [2]. This problem arises in consideration of boundary value problems for mixed type equations [9] and in the theory of selfadjoint operator pencils [11].
Apparently, the first papers devoted to this problem were [6, 1]. We also mention the papers [7, 8, 10].
The aim of the first part of this paper is to generalize some of the sufficient conditions from [1,10]. In the second part, we shall apply these results to some boundary value problems for mixed type second order operator-differential equa-
[373]
tions. There are numerous papers devoted to such equations. In connection with our results we refer to [3, 4] where the existence and uniqueness of generalized solutions of a boundary value problem was proved. In this paper we shall prove the existence and uniqueness of solutions for two boundary value problems (the second is actually dual to the first). One of these problems was considered in [3, 4]. We shall also investigate the question of the smoothness of solutions in dependence on that of the data. This problem is not so simple as in the case of hyperbolic or elliptic equations.
1. Preliminary results. Let E+, E−, E0 be the spectral projectors of B corresponding to the positive and negative parts of the spectrum of B and to ker B. Thus if U = E+ − E− then U B = |B| = BU . The operator U is an isomorphism of F0 onto F0 and U2 = I. We define F−1 to be the completion of F0 with respect to the norm
kukF = kL−1BukF = kBukH0
where H0 is the completion of E with respect to the norm kϕkH0 = sup
ψ∈H1
|(ϕ, ψ)|
kψkH1 .
Throughout this paper ( , ) denotes the inner product in E. An element u ∈ F1
is an eigenfunction of the problem (1) if the equality (1) holds in H0 for some λ 6= 0. Under our hypotheses the operator L−1B is bounded from F1 to F1 and we assume that it is completely continuous. Let {ϕ±i } be the eigenfunctions of the problem (1) corresponding to the positive (λ+i ) and negative (λ−i ) eigenvalues.
We assume that {λ+i }, {−λ−i } are nondecreasing sequences.
R e m a r k 1. We note that if ker B ∩ H1= {0} then F1 coincides with H1. Moreover, the case F16= H1can be reduced to the case F1= H1. In fact, we take the orthogonal projector P in H1on F1and define B1= BP + (I − P∗)B1(I − P ), where B1is any selfadjoint operator such that H1⊂ D(|B1|1/2) and the condition B1v = 0 (v ∈ ker B ∩ H1) implies that v = 0; P∗ is the E-adjoint to P . Then B1v = Bv for any v ∈ F1, F1 is an invariant subspace of L−1B1: H1 → H1 and ker B ∩ H1= {0}.
For convenience of the reader we now present some results from [10]. The next four theorems can be found there. Let L(H, H) be the set of bounded linear operators from H to H.
Theorem 1. The eigenfunctions of the problem (1) normalized by kϕ±i kF0= 1 constitute a Riesz basis in F0 if and only if [F1, F−1]1/2 = F0 (here [F1, F−1]1/2
is defined by the complex interpolation method [5, 12]).
Theorem 2. Let Fs = [F1, F0]1−s. Suppose that there exists s > 0 such that U ∈ L(Fs, Fs). Then [F1, F−1]1/2= F0.
Let M = E+F , N = E−F .
Theorem 3. Suppose that in F0 there exists a projector P on M or N such that for some s > 0, P, P∗∈ L(Fs, Fs) ∩ L(F0, F0) (P∗ is the F0-adjoint ). Then there exists s0> 0 such that U, E+, E− ∈ L(Fs0, Fs0).
Theorem 3 was formulated in [10] as Lemma 2.3 for s = 1. It is easy to see that for s < 1 the proof is the same.
In the sequel we assume that (Bϕ±i , ϕ±j ) = ±δij. We define P±f = ±X
(Bf, ϕ±i )ϕ±i .
If [F1, F−1]1/2 = F0 then any f ∈ F0 can be represented in the form f = P∞
i=1[(Bf, ϕ+i)ϕ+i − (Bf, ϕ−i )ϕ−i ] = (P+ + P−)f . This series converges in F0, and we can introduce an equivalent norm in F0 by setting (see [10])
kf k20=
∞
X
i=1
[ |(Bf, ϕ+i )|2+ |(Bf, ϕ−i )2| ].
Define F0± = {u ∈ F0: P±u = u}.
Theorem 4. If [F1, F−1]1/2= F0 then there exist P0± ∈ L(F0, F0±) such that E±(v − P0±v) = 0 (∀v ∈ F0), P0±v = v, P0±E±v = v (∀v ∈ F0±).
Now we shall give some new sufficient conditions which imply that [F1, F−1]1/2
= F0. They extend the conditions used in [1]. First we list some hypotheses.
1. There exists a space H1⊂ D(|B|1/2) and linear operators X ∈ L(H1, H1), Y ∈ L(H1, H1) such that for some c > 0
(a) kE+ϕkF0 ≤ ckXϕkF0 (∀ϕ ∈ H1),
(b) (|B|Xϕ, ψ) = (Bϕ, Y ψ) (∀ϕ ∈ H1, ∀ψ ∈ H1).
2. There exists a space H1⊂ D(|B|1/2) and linear operators X, Y ∈ L(H1, H1) such that
(a) kE+ϕk ≤ ckXϕk (∀ϕ ∈ H1),
(b) (|B|Xϕ, ψ) = (Bϕ, Y ψ) (∀ϕ, ψ ∈ H1), (c) if F1= (H1∩ ker B)⊥ then for some s > 0
Fs = (F1, F0)1−s= (F1, F0)1−s.
3. There exist a space H1 ⊂ D(|B|1/2), H1 ⊂ H1 and linear operators P, X, Y ∈ L(H1, H1) such that P + Y X ∈ L(H1, H1), ∆P ⊂ H1∩ ker B (∆P being the range of P ) and
(a) kE+ϕkF0 ≤ ckXϕkF0 (∀ϕ ∈ H1), (b) (|B|Xϕ, ψ) = (Bϕ, Y ψ) (∀ϕ, ψ ∈ H1).
Theorem 5. Under any of these hypotheses [F1, F−1]1/2 = F0.
P r o o f. The proof in all cases is almost the same. For example, assume hy- pothesis 3. We define
W = {u ∈ L2(0, ∞; F1) : du/dt ∈ L2(0, ∞; F−1)}.
Then a ∈ [F1, F−1]1/2 if and only if there exists u ∈ W such that u(0) = a (see [5]) and we can set
kak[F1,F−1]1/2 = inf
u(0)=a
kukW = inf
kuk2L
2(0,∞;F1)+
du dt
2
L2(0,∞;F−1)
1/2
. Let us prove the injection [F1, F−1]1/2 ⊂ F0. From this it already follows that [F1, F−1]1/2 = F0 (see the proof of Theorem 2.2 in [10], or [1]). Let a ∈ [F1, F−1]1/2. There exists u ∈ W such that u(0) = a. There also exists a se- quence un∈ C∞([0, ∞]; F1) [5] such that supp un is compact and kun− ukW → 0 as n → ∞. Let an= un(0) ∈ F1. We have (with B+ = E+B)
(B+an, an) = kE+ank2F
0 ≤ c2kXun(0)k2= −2c2
∞
R
0
Re(BXu0n(t), Xun(t)) dt
= −2c2
∞
R
0
Re(Bu0n, (Y X + P )un) dt ≤ c1ku0nkF−1k(Y X + P )unkF1 ≤ c2kunk2W. On the other hand,
−(Ban, an) = 2
∞
R
0
Re(Bu0n(t), u0n(t)) dt ≤ c3kunk2W. Using these inequalities we get
(B−an, an) ≤ c4kunk2W, B−= E−B.
Thus, we obtain
kank2F0 ≤ (|B|an, an) ≤ ckunk2W. Hence we deduce that a = u(0) ∈ F0, i.e. [F1, F−1]1/2= F0.
R e m a r k 2. We can use E− instead of E+ in hypotheses 1–3. From now on, we assume that [F1, F−1]1/2= F0. Let C([0, 1]; H) (H a Hilbert space) be the space of continuous functions from [0, 1] to H. Let
kϕk2s =
∞
X
n=1
(|(Bϕ, ϕ+n)|2|λ+n|s+ |(Bϕ, ϕ−n)|2|λ−n|s).
If s > 0 we set Fs = {v ∈ F0 : kvks < ∞} and if s < 0 we define Fs as the completion of F0 with respect to the norm k ks. As already noted, the norms k kF0, k k0 are equivalent. We consider B−1L as an operator from F0to F0. It is defined at least on Lin{ϕ±i } by means of the equality
B−1L X
(c+i ϕ+i + c−i ϕ−i )
=X
(c+i λ+i ϕ+i + c−i λ−i ϕ−i ).
It is easy to see that it is closable. So we can assume that S = B−1L is closed and in this case it is an isomorphism of F2 and F0. Thus, by induction we can introduce in Fs (s an integer, s > 0) equivalent norms
kϕk2Fk =
[k/2]
X
p=0
kSpϕk2Fk−2p.
Also, S is an isomorphism of Fs and Fs−2 with S−1 = L−1B.
2. Boundary value problems for mixed type operator-differential equations. On the interval (0, 1) we consider the equation
(2) Butt− Lu = Bf
with one of the following boundary conditions:
(3) (4)
u(0) = u0, ut(0) = v0,
E−(ut(0) − v0) = 0, E−(u(0) − u0) = 0,
E+(ut(1) − v1) = 0, E+(u(1) − u1) = 0.
As mentioned above we assume that [F1, F−1]1/2= F0.
Theorem 6. Let f ∈ C([0, 1]; F0), u0 ∈ F1, v0, v1∈ F0. Then there exists a unique solution of the problem (2), (3) such that
dku
dtk = u(k)(t) ∈ C([0, 1]; F1−k), k = 0, 1, 2.
P r o o f. We look for a solution of the problem (2), (3) in the form u =
∞
X
k=1
u+k(t)ϕ+k +
∞
X
k=1
u−k(t)ϕ−k = P+u + P−u.
Then u satisfies the boundary conditions
(5) P+u(0) = u0, P+ut(1) = u1, P−u(0) = v0, P−ut(0) = v1. Each u±k satisfies the equation
(6) u±ktt− λ±ku±k = fk±(t) = ±(Bf, ϕ±k) and the boundary conditions
(7) u+k(0) = u0k= (Bu0, ϕ+k), u−k(0) = vk0= −(Bv0, ϕ−k),
u+kt(1) = u1k= (Bu1, ϕ+k), u−kt(0) = −(Bv1, ϕ−k) = v1k. From (6), (7) we get
(8) u+k(0) = c1kexp(
q
λ+k(t − 1)) + c2kexp(−
q λ+kt)
−
t
R
0
exp(−
q
λ+k(t − ξ)) 2
q λ+k
fk+(ξ) dξ −
1
R
t
exp(
q
λ+k(t − ξ)) 2
q λ+k
fk+(ξ) dξ,
c1k = u0k 2 ch(
q λ+k)
+ u1kexp(
q λ+k) 2
q λ+k ch(
q λ+k)
−
1
R
0
sh(
q λ+kξ) 2
q λ+k ch(
q λ+k)
fk+(ξ) dξ,
c2k =
u0kexp(
q λ+k) 2 ch(
q λ+k)
− u1k
2 q
λ+k ch(
q λ+k)
+
1
R
0
ch(
q
λ+k(1 − ξ)) 2
q λ+k ch(
q λ+k)
fk+(ξ) dξ,
(9) u−k(t) = vk0cos(
q
|λ−k|t) + v1ksin(
q
|λ−k|t) +
t
R
0
sin(
q
|λ−k|(t − ξ)) q
|λ−k|
fk−(ξ) dξ.
From (3), (8), (9) it follows that u0= P+u0, v0= P−u0, and
(10) E−(v1+ B1u1) = E−(v0+ g1), E+(u1+ B2v1) = E+(v1+ g2), where
B1u1=
∞
X
k=1
ϕ+ku1k/ ch(
q
λ+k), B2v1=
∞
X
k=1
ϕ−kv1kcos(
q
|λ−k|),
(Bg1, ϕ−k) = 0, (Bg1, ϕ+k) = u0ksh(
q λ+k) ch(
q λ+k)
+
1
R
0
ch(
q
λ+k(ξ − 1)) ch(
q λ+k)
fk+(ξ) dξ,
(Bg2, ϕ+k) = 0, (Bg2, ϕ−k) = vk0 q
|λ−k| sin(
q
|λ−k|)
−
1
R
0
cos(
q
|λ−k|(1 − ξ))fk−(ξ) dξ.
By the definition of the norm in F0 we find that g1, g2∈ F0.
We now show that there is at most one solution of the system (10) in F0. There exist α, β ∈ F0 such that E−α = 0, E+β = 0 and
v1+ B1u1= α, u1+ B2v1= β.
Then v1= P−α, u1= P+β and P+α = B1P+β, P−β = B2P−α. Hence (11) kP+αk20+ kP−βk20≤ kB1P+βk20+ kB2P−αk20≤ kP+βk20+ kP−αk20. On the other hand, (Bα, α) = kP+αk20+kP−αk20, −(Bβ, β) = kP−βk20−kP+βk20. Thus from (11) it follows that (Bα, α) − (Bβ, β) ≤ 0, i.e. α = β = 0. From the second equation in (10) we get
u1+ P0+E+B2v1= P0+E+(v1+ g2).
The first equation in (10) yields
(12) E−(v1− B1P0+E+B2P0−E−v1) = E−(v0+ g1) − E−B1P0+(E+(v1+ g2)).
The second equation in (10) can be rewritten in the form E+u1+ E+B2P0−E−v1= E+(v1+ g2).
So setting α = E−v1, β = E+u1 we obtain an equivalent system (13) α − B−α = g−, β + B+α = g+.
By uniqueness of solutions for (10), ker(I − B−) = {0} (B− : F0→ F0). On the other hand, λ+i → ∞ as i → ∞ (L−1B is completely continuous). This implies that B1and also B− are completely continuous as operators from F0in F0. Thus, a solution of systems (13) and (10) exists and v1, u1∈ F0. Using the representation (8), (9) of the solution u(t) we can easily check that u(k)(t) ∈ C([0, 1]; F1−k).
Theorem 7. Let f ∈ C([0, 1]; F−1), v0∈ F−1, u0, u1∈ F0. Then there exists a unique solution of the problem (2), (4) such that u(k) ∈ C([0, 1]; F−k), k = 0, 1, 2.
The proof is analogous to that of Theorem 6. We get here for P−u(0), P+u(1) a system analogous to (10).
The question arises of the smoothness of the solutions in dependence on the smoothness of the data. Generally speaking, the smoothness of the solutions does not increase with the increase of the smoothness of the data. We need some orthogonality conditions if we want to say something about the smoothness of the solutions. These can be formulated in terms of the eigenfunctions of the problem (1) [8]. However, it seems simpler to formulate them in terms of some special solutions of the adjoint problem. We define
J3(u0, v0, v1, f, g) = − (Sku0, S−1gt(1))F1+ (BSkv1, g(0))
− (BSkv0, g(1)) −
1
R
0
(BSkf (t), g(1 − t)) dt, J4(v0, u0, u1, f, g) = (BSku1, gt(1)) − (BSku0, gt(0))
− (Sk−1v0, g(1))F1−
1
R
0
(Sk−1f (t), g(1 − t))F1dt, Ak = {(g0, g1) : g0, g1∈ D(Sk), E+Skg1= 0, E−Skg0= 0}.
Theorem 8. Let u0 ∈ F2k+1 and f ∈ C([0, 1]; F2k) (k = 0, 1, 2, . . .). Then there exists a solution u of the problem (2), (3) with u(n) ∈ C([0, 1]; F2k+1−n) (n = 0, 1, 2) if and only if for some v0, v1 ∈ F2k such that E−(v0− v0) = 0, E+(v1− v0) = 0 we have
(14) J3(u0, v0, v1, f, g) = 0
for any solution g of the problem (2), (4) with data v0 = 0, f ≡ 0, (u0, u1) ∈ Ak. If such a solution exists and in addition dnf /dtn ∈ C([0, 1]; F2k+1−n) (n = 1, . . . , 2k + 1) then u(n) ∈ C([0, 1]; F2k+1−n) (n = 1, . . . , 2k + 1).
R e m a r k 3. If gi (i = 1, 2) are the solutions of the problem (2), (4) with data v0i = 0, fi≡ 0, ui0, ui1and E−(u10− u20) = 0, E+(u11− u21) = 0 then g1≡ g2. This is a consequence of Theorem 7.
R e m a r k 4. Suppose that the operator L has the property that if E±g = 0 and g ∈ D(S) then E±Sg = 0. In this case if (14) is valid for some v0, v1 then it holds for any v0, v1 ∈ D(Sk) such that E−(v0− v0) = 0, E+(v1− v1) = 0. This means that the orthogonality conditions (14) do not depend on E+v0, E−v1.
P r o o f. Suppose that (14) holds for some v0, v1. Let v be a solution of the prob- lem (2), (3) with v(0) = u0= Sku0, E−(vt(0)−Skv0) = 0, E+(vt(1) − Skv1) = 0, f = Skf . We set u = S−kv. Then u(n) ∈ C([0, 1]; F2k+1−n) (n = 0, 1, 2). Let
uj(t) =
j
X
k=1
[(Bu(t), ϕ+k)ϕ+k − (Bu(t), ϕ−k)ϕ−k].
Then uj ∈ C2([0, 1]; F2k+2) and
ku(n)j − u(n)kC([0,1];F2k+1−n)+ kSk(f − fj)kC([0,1];F0)→ 0 as j → ∞, where fj = ujtt− Suj. Integrating by parts in
1
R
0
(B(Skujtt− Sk+1uj), g(1 − t)) dt =
1
R
0
(Skfj, g(1 − t)) dt,
where g is a solution of the problem (2), (4) with the data indicated earlier, we get
J3(u0j, v0, v1, fj, g) + (BSk(ujt(1) − v1), g(0)) − (BSk(ujt(0) − v0), g(1)) = 0 (u0j = uj(0)). Letting j → ∞ gives
(15) J3(u0, v0, v1, f, g) + (BSk(ut(1) − v1), g(0)) − (BSk(ut(0) − v0), g(1)) = 0.
From (14) and from E+Sk(ut(1) − v1) = 0, E−Sk(ut(0) − v0) = 0 it follows that (16) (B(ut(1) − v1), Skg0) − (B(ut(0) − v0), Skg1) = 0.
From the definition of Akand (16) we obtain E+(ut(1)−v1) = 0, E−(ut(0)−v0) = 0. So u is a smooth solution of the problem (2), (3).
Suppose now that u is a solution of the problem (2), (3) of class mentioned in the theorem. Then repeating the previous arguments we get (15) where v0= ut(0), v1= ut(1) and hence (14) holds. The last assertion of the theorem follows directly from (2).
Analogous arguments are used in the proof of the next theorem.
Theorem 9. Let f ∈ C([0, 1]; F2k−1), v1 ∈ F2k−1. Then a solution u of the problem (2), (4) with u(n) ∈ C([0, 1]; F2k−n) (n = 0, 1, 2) exists if and only if for some u0, u1∈ F2k such that E−(u0− u0) = 0, E+(u1− u1) = 0 we have
(17) J4(v0, u0, u1, f, g) = 0
for any solution g of the problem (2), (3) with data f ≡ 0, u0 = 0, v0 = g0, v1 = g1, (g0, g1) ∈ Ak. If we have (17) and in addition f(n) ∈ C([0, 1]; F2k−n−2) (n = 1, . . . , 2k) then u(n) ∈ C([0, 1]; F2k−n) (n = 0, 1, . . . , 2k + 2).
3. Examples. Here we do not give examples of application of Theorems 1–3 and 5. These theorems are applicable to a wide class of differential operators L if B is the operator of multiplication by some function g changing sign in the domain where L is defined. Numerous examples can be found in [1, 7, 8, 10].
Tricomi’s equation is a well-known example. In the rectangle D = [0, 1] × [−1, 1]
we consider the equation
(18) xuyy+ uxx= f
with boundary conditions
(19) u(x, 0) = 0 (x ∈ (−1, 1)), uy(x, 0) = v0(x) (x < 0),
uy(x, 1) = v1(x) (x > 0), u(−1, y) = u(1, y) = 0 (y ∈ (−1, 1)).
We consider the boundary value problem (2), (3). Here L = d2/dx2 and D(L) = W22(−1, 1) ∩ W◦12(−1, 1) (the definitions of the function spaces used here can be found in [12]). The space E coincides with L2(−1, 1), F1= W◦12(−1, 1), and F0
is the space of measurable functions with finite norm kuk2F0 = R1
−1|x||u|2dx.
Theorems 1–3 and 5 are applicable here [10, 1] and the eigenfunctions of the problem (1) constitute a Riesz basis in F0. Fk is the set of all u ∈ F0 with
1
R
−1 [k/2]
X
m=0
1 x
∂2
∂x2
m
u
2
dx +
1 x
∂2
∂x2
[k/2]
u Fk−[k/2]
< ∞.
In our case A2k = {(g0, g1) ∈ F2k : x1∂x∂22
k
g0= 0 almost everywhere on (−1, 0),
1 x
∂2
∂x2
k
g1= 0 almost everywhere on (0, 1)}. There are only k functions g0linearly independent on (−1, 0) and k functions g1linearly independent on (0, 1) such that (g0, g1) ∈ A2k. On these intervals g0, g1 are some polynomials. By Remarks 3, 4 we can reformulate Theorem 8 as follows.
Theorem 10. Let f (x, y) ∈ C([0, 1]; F2k+1), u0 ∈ F2k+1, and suppose v0, v1
coincide on (−1, 0), (0, 1) respectively with some v0, v1∈ F2k. Then a solution u of the problem (18), (19) with
∂n
∂ynu(x, y) ∈ C([0, 1]; F2k+1−n) (n = 0, 1, 2)
exists if and only if the data of the problem satisfy the 2k orthogonality conditions (here (f, g) =R1
−1f (x)g(x) dx)
−
x 1
x
∂2
∂x2
k
u0, gt(1)
+
x 1
x
∂2
∂x2
k
v1, g(0)
−
x 1
x
∂2
∂x2
k
v0, g(1)
−
1
R
0
x 1
x
∂2
∂x2
k
f (x, y), g(1 − y, x)
dy = 0 for any solution g(x, y) of the equation (18) with data gy(x, 0) = 0, f = 0, g(−1, y) = g(1, y) = 0 ((x, y) ∈ D), g(0, x) = g0(x) (x < 0), g(1, x) = g1(x)
(x > 0), where g0, g1 belong to the above-mentioned function classes. If in ad- dition ∂nf /∂yn ∈ C([0, 1]; F2k−n+1) (n = 1, 2, . . . , 2k − 1) then ∂nu/∂yn ∈ C([0, 1]; F2k−n+1) (n = 1, 2, . . . , 2k + 1).
R e m a r k 5. As shown by a number of examples, the spaces Fs introduced here are most suitable for investigation of various boundary value problems where the spectral problem (1) arises.
References
[1] R. B e a l s, Indefinite Sturm–Liouville problems and half-range completeness, J. Differential Equations 56 (1985), 391–407.
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[3] N. V. K i s l o v, Inhomogeneous boundary value problem for a second order operator-differ- ential equation, Dokl. Akad. Nauk SSSR 280 (1985), 1055–1058 (in Russian); English transl. in Soviet Math. Dokl.
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[7] —, On the solvability of a boundary value problem for a parabolic equation with changing time direction, Soviet Math. Dokl. 32 (1985), 895–897.
[8] —, Properties of eigenfunctions of a spectral problem and their applications, in: Some Applications of Functional Analysis to Problems of Mathematical Physics, Institute of Mathematics, Novosibirsk 1986, 65–85 (in Russian).
[9] —, Solvability of boundary value problems for second order mixed type equations, in: Non- classical Partial Differential Equations, Institute of Mathematics, Novosibirsk 1988, 77–90 (in Russian).
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